Thread: /* Input */ codes stops working when I modularize the code?

Example code 1 - [ OK ]
3Wah7S4K - Pastebin.com

But when I try to modularize the /* Input */ codes then I get all sorts of warnings and errors. I managed to fix most of them but get stuck here where it complains about my n variable.

Example code 2 - [ Error ]
v4HKi1ae - Pastebin.com

This is not the only warning but it's the first warning at this point.

Code:

   C:\Test_Array\example_code_02.c In function `input': 
90 C:\Test_Array\example_code_02.c [Warning] passing arg 1 of `calloc' makes integer from pointer without a cast

Code:

void input(double *array, double *array_copy, int *n)
{
     int i;
 //    double * array = NULL;
 //    double * array_copy = NULL;
 
    printf("How many numbers do you want to input? --> "); scanf("%d",&n);
    // calloc = Memory allocation for array.

    array = calloc(n, sizeof *array);
    array_copy = calloc(n, sizeof *array_copy);

n is a pointer variable, so you must dereference the pointer variable to get the value to which it's pointing:

Code:

array = calloc(*n, sizeof *array);
array_copy = calloc(*n, sizeof *array_copy);

If you've got errors, then post the errors...we're not clairvoyant.

Thanks for pointing me in the right direction. I googled and got help from Binky Pointer Fun Video along with the reading material Pointer Basics.
Now I finally grasp the concept of pointers I think, thanks to the author differentiating and terming the words Pointers & Pointees.

Here's the simplest pointer code I could find. Example code 3 works but when I try to modularize the same code into its own function in Example code 4 then it no longer works.
Can somebody show me how to do this the correct way?


Example code 3 - [ OK ]

Code:

#include <stdio.h>
#include <stdlib.h>

void input(int *, int *);


//int main(int argc, char *argv[])
int main(void)
{
// 1
	int* x;
	int* y;


// 2
	x = malloc(sizeof(int));
	y = malloc(sizeof(int));


// 3
    input(x, y);


// 4
	*x = 1;
	*y = 3;


// 5
//	x = y;


// 6
    printf("x = %d\n", *x);
    printf("y = %d\n\n", *y);

    system("PAUSE");	
    return 0;
}

/*************
* Functions
*************/
/* ___________________________________________________________________________*/
/* Input */
void input(int* x, int* y)
{



}
/* ___________________________________________________________________________*/

Example code 4 - [ Error ]

Code:

#include <stdio.h>
#include <stdlib.h>

void input(int *, int *);


//int main(int argc, char *argv[])
int main(void)
{
// 1
	int* x;
	int* y;


// 2
//	x = malloc(sizeof(int));
//	y = malloc(sizeof(int));


// 3
    input(x, y);


// 4
//	*x = 1;
//	*y = 3;


// 5
//	x = y;


// 6
    printf("x = %d\n", *x);
    printf("y = %d\n\n", *y);

    system("PAUSE");	
    return 0;
}

/*************
* Functions
*************/
/* ___________________________________________________________________________*/
/* Input */
void input(int* x, int* y)
{
// 1
//	int* x;
//	int* y;


// 2
	x = malloc(sizeof(int));
	y = malloc(sizeof(int));


// 4
	*x = 1;
	*y = 3;


// 5


}



/* ___________________________________________________________________________*/

Thanks laserlight that code looks cool.
Could somebody also show me how the same code, should be solved using the dynamic-memory-allocation-technique?

These two ways of dealing with Pointers & Pointees confuses the heck out of me. My teacher mixes these two kind of coding styles without explaining when he uses which in his newbie materials. So I have a hard time understanding this subject, when tutorials I google does the same mistake also.


source: Pointer Basics

Another way to play with pointers in C (or C++) is using the ampersand (&) operator to compute a pointer to local memory in the stack. However, pointees dynamically allocated in the heap are the most common, so that's what we show.

Static memory allocation:

int x[100]; /* this memory is allocated on the stack */

Dynamic memory allocation:

int *x;

x = malloc(sizeof(int) * 100); /* this memory is allocated on the heap */